Problem: The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $22$ years; the standard deviation is $2.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living between $26.8$ and $29.2$ years.
Solution: $22$ $19.6$ $24.4$ $17.2$ $26.8$ $14.8$ $29.2$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $22$ years. We know the standard deviation is $2.4$ years, so one standard deviation below the mean is $19.6$ years and one standard deviation above the mean is $24.4$ years. Two standard deviations below the mean is $17.2$ years and two standard deviations above the mean is $26.8$ years. Three standard deviations below the mean is $14.8$ years and three standard deviations above the mean is $29.2$ years. We are interested in the probability of a porcupine living between $26.8$ and $29.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the porcupines will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the porcupines will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of porcupines between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular porcupine living between $26.8$ and $29.2$ years is $\color{orange}{2.35\%}$.